precalculus

[viscid thistle]

I say f'(x)

oops wrong lmao

saw*

🤔

oh well i just wrote that instead of saying m (of some equation)

well, it looks like calculus, so, yeah, I mean, I usually hang out on this channel the most, so yeah

¯_(ツ)_/¯

[naive hearth]

I'm in class called precalculus, we have our first exam tomorrow and I know a question like this will be on the test.

[viscid thistle]

ok so you got to this part then plore

=tex y - \frac{7}{8} = \frac{4}{3}(x+(\frac{2}{3})

[viscid thistle]

I took precalc, I never had derivatives ;~;

@naive hearth did u get to that step above?

[naive hearth]

Yeah! I'm working on it now, thanks again this is a big help.

I'm going to try using the distributive property, then adding 7/8 to both sides...

now I have
=tex y - \frac{7}{8} = \frac{4}{3}x + \frac{8}{9})

y -7/8 = 4/3x + 8/9

now adding 7/8 to both sides

which gives me y = 4/3x + 127/72

But, I think the answer i'm looking for is y = 4/3x - 11/6...

[viscid thistle]

im in class rn ill look it over after and post here/dm u sorry

[bitter mist]

Identify the vertex and x-intercepts. Use this information to sketch the graph of each function. f(x)= x^2 + 6x + 3

[naive hearth]

Whoohoo! I got it right! y=4/3x +127/72 passes through (-2/7, 7/8)

Thanks everyone! I'm feeling like I know what I need to work on to get better at this now. 💪 😄

[bitter mist]

Can I get some help working out the problem I posted? I'm new to the server I'm not sure if I am asking in the right area or not sorry.

[twilit beacon]

hello i need help maybe please?

how do i solve this with matrices?

[dusk kettle]

what have you done so far? the first step should be obvious if you know how to do this at all

[twilit beacon]

i dont, thats all my teacher gave me cause i missed class today

i need the inverse or something

i give up though theres no way i'll be able to get it

[dusk kettle]

do you know how to write it as an augmented matrix?

[twilit beacon]

no

[dusk kettle]

oh boy

[twilit beacon]

i've only done systems of 2 equations

[dusk kettle]

like 2 equations with 2 variables?

[twilit beacon]

yeah

[dusk kettle]

do you know how to write those as an augmented matrix?

[twilit beacon]

no idea what this is

thanks for the help but i have to go to bed now

[dusk kettle]

https://en.wikipedia.org/wiki/Augmented_matrix

particularly scroll down to where it has a bunch of equations represented as a matrix

[twilit beacon]

thanks man

[viscid thistle]

@Szechuan Sauce#3229

[wicked epoch]

hey, i missed a lesson and im dying. i'm asked to solve √(x) + 5 = √(2x + 1) algebraically. this is in "solving radical equations using factoring". I've been trying to get this but i feel im missing something disgustingly obvious.

[viscid thistle]

Oh boi, one of those problems

Imma let the next guy handle it, Im off to bed

sorry

[wicked epoch]

no prob

[viscid thistle]

heya koala

[wicked epoch]

heyo

[viscid thistle]

lets work on that nasty thing, but first lets rewrite the radicals as exponents

=tex x^{\frac{1}{2}} +5 = (2x+1)^{\frac{1}{2}}

[viscid thistle]

(I'm doing it on paper to make sure, gimme a sec xd)

[naive hearth]

btw, that link you shared on perpendicular lines was helpful, thanks again @viscid thistle

[viscid thistle]

aight got it @wicked epoch

[wicked epoch]

cool

[viscid thistle]

let me write it down in the latex thing

[naive hearth]

I'm on the edge of my seat waiting to see how the (2x +1)^1/2 gets expanded...😮

[viscid thistle]

there it is @wicked epoch

[wicked epoch]

in all it's glory

[viscid thistle]

fkin latex takes me ages to write lol

lmk if there's anything u dont understand

the bare essential behind this is the idea that

=tex (a^\frac{1}{x})^x = a^{1} = a

[wicked epoch]

i got all the way to 100x = x^2 + 48x + 576, but didnt think to set it to zero

thanks dude, really appreciate it

[viscid thistle]

np

i was kinda iffy on it too, was a good practice

also watch your signs it was -48x

[wicked epoch]

right

[viscid thistle]

=tex (a\pm b)^2 = a^2 \pm 2ab + b^2

[viscid thistle]

(the outer terms, i.e a and b, are always positive since even if they were negative, a negative number ^2 = |number ^2|

(negative * negative = positive)

[hallow ember]

I don't understand how to solve this:

halp

pls

I got a test on thursday

aiming to get 100

[tired nest]

@hallow ember Imagine a y = 2 line in the x-y graph. What happens to a line when you reflect it over y = 2? You can try drawing a picture to see how it works. Which point(s) will be the same between the give (x) and the reflected f(x)? How does the gradient change? How do the x and y intercepts change?

[hallow ember]

hmmm

[tired nest]

And think about the fact that reflection of g(x) about the y = 0 axis is f(x) = -g(x).

[hallow ember]

ooo

ok

I see now

[tired nest]

That's great. 😄

[hallow ember]

I got

[tired nest]

👌

So you can see now, with y = k, you have the negative gradient, then you just have to figure out the change in the constant part.

[hallow ember]

I get it now! Thanks @tired nest 👍

[cold saddle]

hi

is precalculus hs level math?

[willow bear]

it refers to the class called "precalculus" in the US curriculum

usually involves things like trig, rational functions, maybe exp and logs as well

that sort of stuff

[cold saddle]

Yeah that's hs then 😃

[late haven]

oughta put that in the topic

US centricism makes me wanna go waterboarding at Guantanamo

[willow bear]

sorry

i mean i'm not even American lol

[late haven]

lol ik

[viscid thistle]

oi, no politics >:C

[late haven]

oh, right. sorry, i forgot the standard for 'politics' was lowered recently

heh

[willow bear]

yo

[viscid thistle]

lol

[willow bear]

can the shit-flinging not start 👀

[viscid thistle]

^

No politics, this is math.

( but we can apply math to politics 👌 )

[patent beacon]

I'm Canadian. It's called "advanced functions" here. Teaches the harder half of precalc

[viscid thistle]

harder half of precalc? 🤔

[patent beacon]

It's like polynomials of degree higher than 2, radian measure, logarithms

[viscid thistle]

yep, we had that over here in the us as well

[patent beacon]

Might also just be precalc. Never took American precalc

[viscid thistle]

ah, its not that bad

that stuff really isnt too bad

its fun

precalc us had some review on algebra and properties, then review stuff like pythagorean theorem and midpoint formula, and then the "precalc" curriculum began

was like rational functions, quadratics, synth division, function transformation/translation

and then trig ofc

pythagorean theorem, god bless it

lol

haha

ye

[viscid thistle]

quadratics is algebra

i think

[outer brook]

""algebra"" means a lot of things

[twilit beacon]

sure does.

[viscid thistle]

im learning it in algebra 1 this year

so yeah

[signal mesa]

i can see that its negative infinity but is there a way to make it obvious with algebra

[willow bear]

when x approaches 2, 1-x < 0 while (x-2)^2 is positive and close to zero

[signal mesa]

so i can't rationalize the expression at all

[willow bear]

what do you mean rationalize

it's already a rational expression

[signal mesa]

o

change it from an indeterminate form

[willow bear]

it's not an indeterminate form either

[viscid thistle]

@signal mesa you can use partial fractions

=tex \frac{1-x}{(x-2)^2} = -\frac{1}{x-2}-\frac{1}{(x-2)^2}

[signal mesa]

hmm

[viscid thistle]

hopefully this is more obvious for you

to show that the limit is -infinity

[signal mesa]

thank-you

[hexed shard]

hello my fellow nerds

[tired nest]

Hello.

[inner mountain]

A question: - if slope of a vector tangent to a curve (x^3)/2+1/2 at (1,1) is 3/2 how to write that vector equation?

In I,j,k form ?

[dusk kettle]

(1,3/2)?

i + 3j/2?

Plus the point

[austere bramble]

you have no k-th component in R^2

[dusk kettle]

i + j + t(i+3j/2) forall t

[ebon spoke]

<@&286206848099549185>

[viscid thistle]

u know what coterminal angles are

?

[ebon spoke]

When the two terminal sides are the same

How would i do that with radians

[viscid thistle]

well same initial/terminal sides

do you understand how to do it with degrees*?

[ebon spoke]

no

Yeah

[viscid thistle]

ok

how do you do it in degrees?

[ebon spoke]

360+ or - the number right

[viscid thistle]

n360 $$\pm$$

[viscid thistle]

n is an integer

cause u can do like 360, 720, etc

and still find coterminal angles

[ebon spoke]

So

[viscid thistle]

with that in mind

it should be pretty easy to see how it is done in radians

[ebon spoke]

1080 is coterminal with those numbers right

[viscid thistle]

huh

no

those are increments that you'd be making to your angle

like if your angle given was 5 degrees

coterminal angles can be 365, 725, -355, -715

(in degrees for those)

[ebon spoke]

ooo ok ok

Kk lets do radians now

[viscid thistle]

so do you see how it should be in radians now? 😃

[ebon spoke]

Should i convert it to degrees first

[viscid thistle]

no

[ebon spoke]

Then back to radians after

Oh

[viscid thistle]

that's a waste of time xd

[ebon spoke]

But would that work tho

[viscid thistle]

yes but you'd be wasting time

lol

think about it

what are we doing now to get the coterminal angle?

[viscid thistle]

no not that

i mean what is the procedure u follow to get a coterminal angle

if u had to describe it in words

[ebon spoke]

Pretty much adding 360 to an angle right

[austere bramble]

or subtracting

[ebon spoke]

Like how u did 5+360

[viscid thistle]

we are grabbing an integer "n", and multiplying it by "360" then adding/subtracting that to our angle that we want to find a coterminal angle for

[austere bramble]

in this case 2pi since its radians

[viscid thistle]

n has no meaning of itself

but ye, 360 means one revolution

hence the 2pi

i was trying to get u to see that from the logic lmao

[ebon spoke]

Kk

[viscid thistle]

do you understand now?

[ebon spoke]

Lets do 15 as an example and ill see if i understand

[viscid thistle]

why its n2pi

alright

[ebon spoke]

Im ready

[viscid thistle]

do it then

and share the answer

[ebon spoke]

2pi/6?

4pi/6

8pi/6

Are they all coterminal

[austere bramble]

no

you need yo add a seperate 2pi

so if youre angle is 2pi/6=pi/3

then pi/3+2pi

is coterminal

[viscid thistle]

we couldnt go throught it much over the explanation but the reasoning behind the +2pi

is cause u complete a revolution

(aka u get back to that point in the circle)

[ebon spoke]

😐

I know this is right under my nose but im not getting it

[viscid thistle]

aight so

[ebon spoke]

I get the revolution part a bit but

[viscid thistle]

lets go back to the explanation

let me draw sth up

ms paint skills one sec

the drawing was poor, let me find one online

@Michael/Onzurk#6177 still here?

[ebon spoke]

@viscid thistle mb

[viscid thistle]

well u can see feom the drawing above

that we complete 1 (or more) revolution, either in a positive or negative direction, to find a coterminal angle

that revolution in a circle translates to 360 degrees, or 2pi

[split elm]

Okay

[timber maple]

yee

[split elm]

https://cdn.discordapp.com/attachments/326138737606262786/369467301793103873/latex.png

Problem

So

Try to use that sine rule

First

[timber maple]

Yes! I did get till, sin(A) + sin(B)/ sin(C)

[split elm]

Okay

Now

What is sin(A) + sin(B)

?

[timber maple]

i know

wait

2sin(a+b/2)cos(a-b/2)

am i right?

[split elm]

Wait

I didn't remember the formula, and I don't bother to derive

[timber maple]

Oh k :3

[split elm]

But

Yeah

It is

Well

You're close

To proving

Just think what information you didn't use

[timber maple]

OH!!! WAIT!! I JUST GOT IT

[split elm]

Okay good c:

[timber maple]

I was too stupid

[split elm]

Nah

That problem is a bit intimidating

[timber maple]

should have written sinC as 2sinc/2cosc/2 ;-;

[split elm]

Yeah

[timber maple]

Thank you very much :3

I have another problem.... i need help in it too

[split elm]

Post it

[timber maple]

wait... gotta write dem tex ;-;

=tex \text{For any } \Delta \text{ABC, Prove that,} a(b cos C - c cos B) = b^2 - c ^2

[timber maple]

there ;-;

the last one, i knew where to start... this one... i dont even know where to start

[split elm]

Wait a minute

I didn't do trig for a long time

Sorry

[timber maple]

Alright! Thank you very much tho :3

You helped me out a bunch

[split elm]

Wait I'm still figuring out

@timber maple try to use law of cosines

I feel that

Like a^2 = b^2 + c^2 - 2bc cos(A)

Yeah, use law of cosines

My thought process is like : because there's no fraction involved, that's why it's not suggested to use Sine Rule since we can't cancel out 2R, then just random hitting of law of cosines

[trim sigil]

Can anyone confirm the m of angle a on the bottom? I already know I got h wrong and it should have stood out that it couldn't possibly be that large before i submitted this lol

the answer key I got has this area cut off... 😦

[willow bear]

is the 15 supposed to refer to the whole hypotenuse?

[trim sigil]

yes

because I used 15 as the hypotenuse and got theta right

[willow bear]

== asind(8/15)

[granite stirrup]

32.23095264

[austere bramble]

you mean arccos?

[trim sigil]

== arccos(8/15)

[granite stirrup]

Error: Unknown name: arccos

[willow bear]

acos

[trim sigil]

== acos(8/15)

[granite stirrup]

1.00826008

[willow bear]

acosd if you want degrees

[trim sigil]

== acod(8/15)

[granite stirrup]

Error: Unknown name: acod

[trim sigil]

== acosd(8/15)

[granite stirrup]

57.76904736

[willow bear]

also no, @austere bramble arccos(8/15) would be α, not θ

[trim sigil]

ok so a is correct, thanks

[austere bramble]

didn't he ask for alpha?

[trim sigil]

yeah

[willow bear]

57.77

[trim sigil]

couldn't think of that letter just woke up after 4 hours to revise this real fast

yeah 57.77

Thanks!

can anyone suggest an online graphing calculate that shows asymptotes?

[trim sigil]

So this is the graph they provided for [(x-2)/3]-[5/(2y-4)]=0

and this is the graph I got through mathway

wondering why I am missing that horizontal line around y=1.something

[dusk kettle]

that horizontal line is just them being bad at graphing

they connected the +infty and the -infty when they shouldn't have

[trim sigil]

ok I am going to mark that as correct then... I am doing a self evaluation currently

[white bronze]

sin^3(x)+cos^3(x)/ sinx+cosx = 1-sinxcosx It's a trig identity proof. Could someone just get me started with the first step? I'm not sure what to do.

[patent beacon]

@white bronze Factor the top as a sum of cubes

That's really all you'll need to do. The rest should follow

[white bronze]

Thanks @patent beacon I probably should have known to do that.

[patent beacon]

Np at all. Feel free to ask anytime

[viscid thistle]

=tex sin^n (x) = (sin x)^n

[viscid thistle]

i was very confused by that too until taking precalc a second time, so just recognize thats what it means :p

[white bronze]

=tex tan2x =2tan(x)/1-tan^2(x)

[white bronze]

@patent beacon mind helping me with this one too?

2tanx is being divided by 1-tan^2x

[patent beacon]

My immediate reaction is to convert everything to sin and cos, and notice that sin(2x) = sin(x + x). Probably not the fastest solution

[white bronze]

it's another trig identiy proof

so im trying to change only one side

[patent beacon]

You should be allowed to change both sides, no?

[white bronze]

not what my teacher said

[patent beacon]

Welp. That will make it harder

Naturally you'll have to change the left. You can break that into
sin(2x) / cos(2x). There should be some identities to cover that

[white bronze]

for cosx would i use cos^2x -sin^2x? @patent beacon

[patent beacon]

Yeah. And 2sinxcosx for sin.

Divide top and bottom by cos²x and...

[white bronze]

why am i allowed to divide cos^2x?

[patent beacon]

You're allowed to divide both the numerator and denominator by cos²x, since that would be the same as dividing by 1

They cancel eachother out

[white bronze]

oh ok so i'm multiplying by 1?

[patent beacon]

Ya ya pretty much, if you're used to that

[viscid thistle]

sin 2x is equal to 2 cos x sin x not sin (x+x) btw

sorry that im just butting in but thats sth important to note

[patent beacon]

@viscid thistle It's equal to both.

But why? Who said the second one?

Oh I see way above

[viscid thistle]

the double angle formula defines it

[split elm]

It's equal to both

Why it doesn't ?

In fact you got 2cos(x) sin(x) from sin(x+x) = sin(2x)

[patent beacon]

I mean, you're right, not the best way to put it, but I corrected it below.

[viscid thistle]

eh maybe i should keep my mouth shut since its 12:30am but still id use the double angle formula

anyways sry for butting in keep going

[valid oracle]

@white bronze not being able to change both sides is a little silly, but you can effectively do the same thing and combine the steps at the end if you understand what i mean

[gleaming schooner]

Hello

[willow bear]

yeah ok so

any polynomial will work honestly

[gleaming schooner]

Well, I have an example here:

=tex x^5 - 5x^3 + 4x

[willow bear]

mhm

[gleaming schooner]

Use the leading term to determine the end behavior of the graph of the function

Leading term is 1.

Positive.

[willow bear]

no

[gleaming schooner]

x^5 *

Right?

[willow bear]

1x^5 is the leading term

yes

positive, odd degree

so it's clear that when x is large, the x^5 term is all that matters, right?

x^3 and x are both tiny compared to it

[gleaming schooner]

Yea, why is that the case though? Is that because being the highest degree term, it has the most influence over the value of the function?

Or the most impact on the value of the function*

Rather

[willow bear]

x^5 = x^3 * x^2 = x^3 * (HUGE)

and well

yes

[gleaming schooner]

What do you mean * (HUGE)? 😮

I understand that x^5 = x^3 * x^2

[willow bear]

well if x is large

then so are all of its positive powers

[gleaming schooner]

In Sal's videos, he kept mentioning "really really positive" and "really really negative" numbers.

[willow bear]

yeah

[gleaming schooner]

I don't even know what that meansl ol

lol*

How can a number be "really really" negative?

Isn't a negative number a negative number

[willow bear]

like -1000000000 and [phone]

[gleaming schooner]

Ah.

[willow bear]

far away from zero

[gleaming schooner]

So just large numbers.

[willow bear]

precisely

large numbers, distinguished by sign

so anyway, when x is large and positive, then x^5 is also positive

and when x is large and negative, then x^5 is also negative

[gleaming schooner]

Alright, that seems clear enough

And that's because we have a negative exponent, right?

[viscid thistle]

maybe this will help too

https://www.desmos.com/calculator/ovujhdndsw

[willow bear]

@gleaming schooner no, all the experts involved are positive

also

for polynomials, there are really only four different patterns of end behaviour

[viscid thistle]

i'll make a table in word lol

oh

u beat me xd

[gleaming schooner]

Checking out that graph now

So are these patterns that need to be memorized?

I wanted to develop an intuition for this

I'm having trouble wrapping my mind around the concept

Which bothers me

[willow bear]

you don't need to memorize them

numbers keep their sign when raised to odd powers

and become positive when raised to even powers

make sense?

[gleaming schooner]

Yep, I'm just going over your table

3^2 = 9

9 is odd.

2 is even

[willow bear]

i said numbers keep their signs

[gleaming schooner]

Ah

Yes lol

[willow bear]

parity is irrelevant

1000001 and 1000000 are not that much different

[gleaming schooner]

and become positive when raised to even powers

You said they become positive when raised to even powers. Right?

[viscid thistle]

excuse shitty word lol

[willow bear]

any number raised to any even powers becomes positive, yes @gleaming schooner

[gleaming schooner]

What's LC stand for @viscid thistle

[viscid thistle]

leading coefficient

[willow bear]

leading coefficient

[gleaming schooner]

Alright

@willow bear But what about the 3^2 example

3 is odd

[willow bear]

you

[gleaming schooner]

And gets raised to 2, which is even

[willow bear]

don't care

about the parity of the base

[viscid thistle]

you only look at the exponent

[willow bear]

it's the parity of the exponent that matters

[viscid thistle]

i.e the base as ann says of (3) is irrelevant, only the sign of the leading coefficient is important

[gleaming schooner]

Oh

[viscid thistle]

what "3" tells you is that it is a positive lc

[gleaming schooner]

I'm sorry

[viscid thistle]

nothing more xd

[gleaming schooner]

We're talking about the sign

Lol.

So sorry.

[willow bear]

(-100)^2 = 10000
(100)^2 = 10000

both positive

[gleaming schooner]

Yes, I understand.

For some reason I was thinking even/odd. Not about the sign lol

[willow bear]

"-x^{2n}" in my table refers to negative leading coefficient and even degree, just in case

[gleaming schooner]

So, that's basically what I need to understand most importantly, right? Is how degrees affect the parity of the leading coefficient.

[willow bear]

🙃

right words, very wrong order

[gleaming schooner]

How the parity of the degrees affect the signage of the leading coefficients?

[willow bear]

these two things affect end behavior

[gleaming schooner]

Alright

I understand.

[viscid thistle]

👌

[gleaming schooner]

So, now, let me try to apply this to my example

[viscid thistle]

can u repost it pls

[gleaming schooner]

Sure

[viscid thistle]

vicious handwriting ann

calm down xd

[gleaming schooner]

Haha

I'm jsut writing this all down

One moment

just*

I think you mixed it up @willow bear

You said numbers keep their sign when raised to odd powers and become positive when raised to even powers

earlier

And wrote even degree: same signs

[willow bear]

i did

[gleaming schooner]

On your paper

[willow bear]

same signs as in, the polynomial goes to the same sign of infinity in both directions

both positive or both negative, depending on the leading coefficient's sign

[gleaming schooner]

Alright, I'm gonna try a few examples with you guys to make sure I have an understanding.

=tex f(x) = x^5 - 5x^3 + 4x

[gleaming schooner]

So, the leading term is x^5

We've got an odd degree, which means that he leading coefficient will become negative.

[willow bear]

We've got an odd degree
yes
which means that the leading coefficient will become negative.
wat

[gleaming schooner]

D; Umm

Let me go over it, one sec.

Yea

Idk what I'm talking about. Excuse me.

Don't know where my head is atm.

We've got a positive leading coefficient and an odd degree.

The LC keeps its sign.

[willow bear]

The LC keeps its sign
implied "when x goes to positive infinity"

[gleaming schooner]

Yea

I'm gonna graph it on Desmos so I can visualize it

[gleaming schooner]

So:

=tex f(x) = x^5 - 5x^3 + 4x

[gleaming schooner]

Will have the same end behavior as x^3

As x -> ∞, y -> ∞

As x -> -∞, y -> -∞

[willow bear]

x^5, not x^3, but otherwise yes

[gleaming schooner]

Well, I meant that our example function

Will have the same end behavior as the graph of y = x^3

I just mentioned that because I was using the y = x^3 as a starter to visualize everything

So what about turning points?

How do I develop an understanding of that

Number of turning points that the graph of a polynomial has.

Or max turning points.

[willow bear]

turning points?

well, those, if any, are the zeroes of the derivative of your function

so a polynomial of degree n can have no more than n-1

[gleaming schooner]

What are the zeroes of the derivative of the function? What's that mean

[willow bear]

do you know what a derivative is?

[gleaming schooner]

I haven't gotten to derivatives yet.

[willow bear]

mh

[gleaming schooner]

But I'm gonna just keep going through my examples

You don't have to explain those atm

[willow bear]

there isn't really a way to explain turning points without that

[gleaming schooner]

Alright, np.

I've gotta determine the end behavior of the following graph:

=tex g(x) = -2(x-5)(x+1)^3

[gleaming schooner]

I've gotta first distribute all of that, right?

Expand

[willow bear]

well you only really need the leading term

so you can replace x-5 and x+1 with x

so the leading term will be -2x^4

[gleaming schooner]

I graph that

And it's just a parabola

Downwards facing

[willow bear]

well not exactly a parabola

more like a quartic parabola

since yknow, degree 4

[gleaming schooner]

I'm not understanding how to deduce the end behavior on this one

I know that as x -> ∞, y -> -∞

Because we've got a negative LC with an even degree.

But how about when x -> -∞ ?

[willow bear]

well

even degree

you'll get the same sign of infinity in both directions

[gleaming schooner]

I don't get how, a negative number to an even degree will give us a positive number.

[willow bear]

yes, but your leading coefficient is negative

negative * positive = negative

[austere bramble]

if n is an even number, then n=2k for some integer k. Then (-1)^n = (-1)^(2k)=[(-1)^2]^k=1^k=1

[viscid thistle]

for the number of turns ur graph takes i think it was the change in signs of the function right

i seem to vaguely remember sth about it

[willow bear]

the derivative, not the function itself

[viscid thistle]

oh

[gleaming schooner]

=help latex

=tex f(x) = -3x^5 + \sqrt{2}x + 1/2x

[gleaming schooner]

That isn't a polynomial function, right?

Because the degrees of the leading terms aren't decreasing

[willow bear]

is the last term meant to be $$\frac{1}{2}x$$ or $$\frac{1}{2x}$$?

[willow bear]

if it's the former, your function is still a polynomial, albeit written a bit wonkily

also

"leading terms"

no

there is only one leading term

[gleaming schooner]

The former.

[willow bear]

yeah

[gleaming schooner]

So why isn't the following a polynomial function:

[willow bear]

=tex f(x) = -3x^5 + \left(\sqrt{2} + \frac{1}{2}\right)x

[gleaming schooner]

Ah, strange, they have it written in the book

The way I shoiwed you

Actually, I think x should be in the radical

They don't make it very clear

showed*

[willow bear]

=tex \sqrt{2x}?

[gleaming schooner]

Yea

I believe so

[willow bear]

yeah, no, that would disqualify your thing from being a polynomial

[gleaming schooner]

It doesn't look to be that way.

[willow bear]

polynomials can't contain fractional powers

[gleaming schooner]

How about:

=tex -3x^5 + 2\sqrt{x} + \frac{2}{x}

[gleaming schooner]

That isn't a polynomial

But why?

Is it because x is in the denominator?

Of the last term

[willow bear]

both due to the 1/x and to the sqrt(x)

neither of these can be in a polynomial

[gleaming schooner]

Why is 1/x invalid?

Also, are 'turning points' the same thing as 'inflection points'?

[willow bear]

Why is 1/x invalid?
because polynomials can only include nonnegative powers of x

[gleaming schooner]

Thanks

What channel should I go to for help with factoring polynomials?

[willow bear]

#prealg-and-algebra ig

[weak quartz]

help XD gotta use y = a|x-h|+k

[viscid thistle]

Let me just uh, break my neck and squint to read the problem here

[weak quartz]

LOL sorry man XD

Actually started dying laughing XD

[primal crater]

Help can't find horizontal asymptote

(3x^2 + 8x + 4) / (x - 1)

or don't know how to find horizontal asymptote at this rate

[tired nest]

What do you know about finding horizontal asymptotes so far, taz?

[primal crater]

I know the 3 rules if the numerator has a higher degree than the denomenator than there isn't one, if they are the same divide the coeffecients and if the denomenator is has a larger degree than it's at 0

[tired nest]

Okay, right, so here we don't have a horizontal asymptote, just a slant asymptote.

Because 2>1

[primal crater]

oh okay, well my teacher told me I didn't have to worry about those types of questions yet so I'll just skip it. Thanks for the help

[tired nest]

No problem.

[viscid thistle]

heyy

[vocal wolf]

Could I get some help with this problem

x^2+x+1 > 0

solve it in terms of intervals

<@&286206848099549185>

[past veldt]

Personally, I'd complete the square.

@vocal wolf

[vocal wolf]

Ok, I will try it

I can't seem to solve this

[past veldt]

What have you tried so far?

[vocal wolf]

First I tried quad formula which didn't work then I tried completing the square too

I did (x^2 +x ) + 1 > 0

then 1/2 * 1 -> (1/2)^2 = 1/4

(x^2 + x + 1/4) + 1 - 1/4 > 0

Is this right so far?

[past veldt]

Yes, that's fine so far

[vocal wolf]

then it's (x+1/2)^2 + (3/4)

[past veldt]

Yes

[vocal wolf]

but then sqrt((x+1/2)^2) = sqrt(-3/4)?

[past veldt]

Woah, not quite.
We have
(x+1/2)^2 + 3/4 > 0
Do you notice anything that's the same for both of the terms on the left-hand-side?

[vocal wolf]

is there something in the denominator?

[past veldt]

Hmm. My hint was bad.
Alright, from (x+1/2)^2 + 3/4 > 0 --> (x+1/2)^2 > -3/4
We don't need to do anything more to it; we can tell what values of x satisfy this from here

[vocal wolf]

How do you tell

[past veldt]

Well what can we say about the sign of square numbers?

[vocal wolf]

always positive

[past veldt]

Yep. So are there any values of x where (x+1/2)^2 isn't > -3/4?

[vocal wolf]

oh, no

[past veldt]

Yep. So for all x.

So it's solved

[vocal wolf]

thank you very much

[past veldt]

No problem

[viscid thistle]

ツ

need help anyone?

[vocal wolf]

are you asking for help or offering help @viscid thistle

[viscid thistle]

this isnt craiglist omegalul

just answer for questions when asked xd

[merry nimbus]

Lol

@viscid thistle where were u

To solve d/dx of sec^2 (x)

#calculus

[viscid thistle]

offering @vocal wolf

[merry nimbus]

Oh

[vocal wolf]

@viscid thistle well if you're still offering, I need help with some trig

Could I get some help with this trig? It says to find the exact trig ratios for the angle whose radian is given. 3pi/4

nvm I got it

[split elm]

@vocal wolf just correcting yourself a bit, but x^2 => 0 for all x , so, it's just not all positive, but also with x = 0 counts too

[vocal wolf]

What do you mean by it's not all positive?

@split elm

[split elm]

I mean , all positive x sure satisfy that

But don't forget x = 0 also satisfy the inequality, well I just say this because of how Complexima wording

In your problem , you won't care when x+ (1/2) = 0

[vocal wolf]

ohhhh ok

So it's more like [0,infinity)

[split elm]

In your case, it is not, but if you're asked how does the sign of squared number?

You should answer like that

[vocal wolf]

ah, ok

Thanks

[viscid thistle]

😄

[white bronze]

=tex 2cscx =secxcscx

[white bronze]

It's a trig identity where I'm only allowed to change one side

any helP?

[patent beacon]

@white bronze It's incorrect.
2csc(x) - sec(x)csc(x) = 0
csc(x) = 0, sec(x) = 2
x = pi/3

[white bronze]

oh wait oops

i wrote it out wrong @patent beacon

=tex 2csc2x=secxcscx

[patent beacon]

@white bronze Convert the left to sin. Apply double angle identity. Convert back

You're gonna facepalm

[white bronze]

wait so it'll turn into

[patent beacon]

Indeed!

[white bronze]

=tex 2*(1/sin2x)

[white bronze]

ahh

i see

[patent beacon]

Yaya

[white bronze]

=tex sin(3x)=3sinx-4sin^3x

[white bronze]

any help please?

[tired nest]

What are you wanting to do here?

[split elm]

If you want to prove that

Try to rewrite sin(3x) as sin(x+2x)

[willow bear]

$$\sin, \cos, \tan, \sec, \csc, \cot$$ all exist

[willow bear]

=tex \sin(3x) = \sin(2x)\cos(x) + \cos(2x) \sin(x)

[willow bear]

@white bronze can you take it from there?

@vocal wolf saying "an angle whose radian is given" is kinda like saying "a rod whose inch is given"

:p

[vocal wolf]

That's what my textbook says lol

[willow bear]

verbatim?!

[vocal wolf]

"find the exact trigonometric ratio for the angle whose radian measure is given" is what it actually says

[willow bear]

aha

radian measure

[hexed ermine]

Guys I just Ace'd my precal exam

[viscid thistle]

👍

[viscid thistle]

ur stunting thanks to this discord thi

tho

remember that

xd

[viscid thistle]

@viscid thistle youre stunting???

w-what?!

[viscid thistle]

stunting as in showing off in front of others

ah

nvm

did i miss a pun or joke

xd

...

Im not sure

👌🏼

[white bronze]

=tex sin2x=2/5

[white bronze]

any help?

also

=tex cosx-4/7=0

[late haven]

@white bronze what have you tried?

[white bronze]

i dont even know

i dont think i'm supposed to use a calculator or what

@late haven

[late haven]

well lets try it with the calculator for now until we know what we're doing

do you know the inverse trig functions?

[white bronze]

what do you mean by that?

by knowing*

[late haven]

do you know what sin^-1 and cos^-1 are?

you may have seen them as "arcsin" and "arccos" or "asin" and "acos"

[white bronze]

well yeah I know what they are

I thought you meant they had identities or something

[late haven]

oh, nah

so, for one of your problems sin2x = 2/5

you can take the inverse sine of both sides

=tex \sin^{-1}(\sin(2x)) = 2x = \sin^{-1}(\frac{2}{5})

[white bronze]

so the number with a decimal is my quadrant one answer?

or that divided by 2

[late haven]

uh?

[white bronze]

about 11.789

[granite stirrup]

0.20575842

[late haven]

=calc asin(2/5)/2

=calc asin(2/5)/2 * 180/pi

[granite stirrup]

11.78908924

[late haven]

yes, 11.789 degrees

what this means is that sin(2 * 11.789) equals 2/5

[white bronze]

ok

thanks

[viscid thistle]

uh

if u wanted degrees

i think u had to use the "d" postfix

as in

==sin(1)

[granite stirrup]

0.84147098

[viscid thistle]

==sind(1)

[granite stirrup]

0.01745241

[late haven]

oh, okie

[viscid thistle]

👌🏼

had that happen to me in a quiz

i was mad af lol

[late haven]

lol

I wish there was a postfix calculator command on the bot

[viscid thistle]

wdym

[late haven]

its a style of writing out a calculation for a computer. i think it's super useful for typing calculations quickly

so instead of sin(2*11.789) one would enter 2 11.789 * sin

[viscid thistle]

oh i see

it may get confusing as to whats the arg to sin

so maybe thats why they did it this way

[white bronze]

@late haven Would i subtract 11.789 by pi or should I not because i remember my teacher saying something about sin^1 only operating in pi and -pi

[late haven]

ah, no, you wouldn't need to

[white bronze]

ok

[late haven]

i'll try explain what she was getting at

sine is a repeating function, i'm sure you're aware. so every 360 degrees it reaches the same value

so sin(2 * 11.789) = 2/5

but so does sin(2 * 11.789 + 360) = 2/5

and sin(2 * 11.789 + 720) = 2/5

[white bronze]

no the instructions are find all solutions when x is between 0 and 2pi

[late haven]

and so on. but sin^-1 will only give you one answer, even though there's an infinite number of answers

[vocal wolf]

could someone teach me how to factor 3 term polynomials where a isn't 1 quickly?

I use quadratic formula everytime but I don't think I'll have time for that on my exam

like for example 2x^2-x-1

[willow bear]

mh

if you can guess a root then you can factor that out

[vocal wolf]

just guess huh

like as in I guess a root is (x-1)? then factor it out

[willow bear]

(x-1) is a factor
1 is a root

but well

yes

idk

i'd use the qf myself

[vocal wolf]

ah ok

Guess I'm pretty fast with it anyway

thanks tho

[viscid thistle]

if u mean quadratics

what i usually would do is see how a "splits up"

then try to get my products of "c" that add up to "b" keeping in mind how i'd have to multiply that value of a

i.e, look at a first then come up with c and b as normal but with a in mind

[limpid quiver]

2x^2-x-1=0
(2x+1)(x-1)=0
x= -1/2 or x= 1

[willow bear]

@timid jacinth

[lofty elm]

What is precalculus

[timid jacinth]

So how do I go about finding an angle where tan9pi/10= tan(x)

[willow bear]

tan(9π/10) = tan(x), surely?

okay so

[timid jacinth]

With sin it was very easy cause I just needed to use symmetry to find a y-coordinate

[willow bear]

the symmetry for tan is around the origin

😛

also

$$\arctan(\tan(u))$$ always equals $$u - \pi t$$ for some integer $$t$$

[willow bear]

so in this respect, arctan(tan(u)) is easier to find than say, arcsin(sin(u))

all you need to do is subtract π from your angle until you land between -π/2 and +π/2

or, of course, if u < -π/2, you'll want to add π instead

[timid jacinth]

@willow bear thanks, that really cleared it up for me

[bitter hound]

help with the first one? I know the equations im supposed to use but my work is sloppy and confused

[burnt vigil]

Take vectors, so dog =
26.2(0i, 2.7j)
cat=
5.3(3.04i, 0j)
Momentum of each
(26.2 x 2.7)j =70.74j
(5.3x3.04)I = 16.112i
Add them together for the owners momentum =
(16.112i,70.74j)
If j is y and i is x you can then think of the vector as a triangle and use trig to resolve the angle
Tan^-1(70.74/16.112)= 77.17Ø
To find velocity reduce the momentum triangle by a factor of the weight to make a velocity triangle then use Pythagoras theorem to work out the hypotenuse which equals the magnitude of direction =
sqrt((70.74/74)^2 + (16.112/74)^2) = 0.98ms^-1

@Shaepai#4788

[hard fiber]

z is inversely proportional to t. If t = 3, then z = 5.

how do I write a formula for this?

[willow bear]

do you know what "z is inversely proportional to t" means?

[hard fiber]

yes

z = k/t

[willow bear]

so

[hard fiber]

but how do I interpret the t = 3 z = 5

[willow bear]

z = k/t

if t = 3, then k/t = 5

can you solve for k?

[hard fiber]

15

so would it be 5k/3t?

[willow bear]

...no

where'd you get that?

[hard fiber]

oh

oh

k is 5

my bad

[willow bear]

no

k is not 5

you solved for k yourself!

you even stated what k was, correctly!

it wasn't 5!

[hard fiber]

oh

mb

what would the formula be then?

[willow bear]

what do you mean by formula?

do you want to write down the relationship between z and t?

[hard fiber]

for z is inversely proportional to t. If t = 3, then z = 5.

yea

[willow bear]

well you already wrote that down

z = k/t

except now you know what k is

[hard fiber]

so it is z = 5k/3t?

[willow bear]

no!

how are you getting that from z = k/t and knowing that k = 15?!

[viscid thistle]

you are plugging in the value, not appending it to the already existing equation

[hard fiber]

so do I just disregard t = 3 and z = 5?

[willow bear]

what? no

you used these to find the value of k

[hard fiber]

this is the question

Express the statement as a formula. Use the given information to find the constant of proportionality.

so is it z = 5/3t?

[willow bear]

sigh

no

you found the constant of proportionality!

you found k!

you said it was 15!

you were right!

and then you proceed to screw it up =.=

[hard fiber]

how do I express it as a formula?

[willow bear]

you said

k = 15

you know that the formula is z = k/t

[hard fiber]

so the formula is z = k/t

[willow bear]

NO!

NO

[hard fiber]

and the constant of proportionality is 5/3

[willow bear]

NO!

[crude kayak]

Would this happen to be a common trig formula?

If so, any chance anyone knows of any resources that have similar identity/formulas like this one?

[patent beacon]

Make a right angled triangle that obeys sin¯¹(z). That must mean opposite = z and hypotenuse = 1. Use pythag for the third side. Now, take the cot of this triangle

[hallow ember]

Hello again. I've been wondering how one would use synthetic division (with the chart) to solve this question. Thanks in advance 😃

[viscid thistle]

🤔

[hallow ember]

🤔

[viscid thistle]

is this more like algebra

[hallow ember]

idk

it's in my precal workbook

[viscid thistle]

k

https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials#synthetic-division-of-polynomials

idk if this helps

[patent beacon]

@hallow ember You can't synthetic out something with an unknown constant. It just won't work

[hallow ember]

hmmm

[patent beacon]

However, difference of cubes will help you here.

[hallow ember]

apparently the answer is

🤔

wow

it was that simple

[viscid thistle]

😶

[hallow ember]

dumb question to put in the book

o well

i get it now

thx

[patent beacon]

Did you understand the difference of cubes hint?

[hallow ember]

Yep

but I don't get why the book wants me to use synthetic division to solve this

[patent beacon]

Wait, they do want synthetic? That is weird

[hallow ember]

Yeah I might have to ask the teacher about it

[willow bear]

1, 0, 0, -a^3 🤷

[hallow ember]

b) is the answer

I just don't know how you get x^2 + ax + a^2 from 1, a, a^2, and 0

[willow bear]

0 is the remainder

1x^2 + ax^1 + a^2x^0

[hallow ember]

O now I see

Thanks

[timid jacinth]

How do I do (x+2)/-x^3?

[willow bear]

... $$ \frac{x+2}{-x^3} = -\frac{1}{x^2} - \frac{2}{x^3}$$?

[willow bear]

@timid jacinth i have no idea what you're trying to do and it'd really help if you showed the problem you need help with exactly as it's stated

[timid jacinth]

Alright one sec

@willow bear https://gyazo.com/0c6066cb498d07700db180d3112633a3

Sorry, I'm on a phone

[twilit zodiac]

is it (x+2) times square root or 2 times square root

[timid jacinth]

It's a long division bar

I'm supposed to do long division

[willow bear]

aha ok

you know what makes sense?

writing out those polynomials in a bit of a formalized fashion

i'll demonstrate what i mean in just a moment

[timid jacinth]

In class the teacher just started dividing this

[willow bear]

bit cleaner isn't it?

[timid jacinth]

Yeah

What are the x2 x1 and x0 up top for

[willow bear]

just to remind you that this is a polynomial :p

[timid jacinth]

Ok

[willow bear]

polynomial long division is even simpler than number long division imo 😛

no carries at all

does what i wrote make sense to you?

...fuck

i dun goofed

[timid jacinth]

Eh

BIT

a biy

Bit*

I just don't see where the + went on x+2

[willow bear]

corrected

also i'm just writing out the coefficients

=tex ax^2 + bx + c = \stackrel{x^2}{a} ; \stackrel{x^1}{b} ; \stackrel{x^0}{c}

[timid jacinth]

Yeah I get lost by this

I don't understand how (x+2) can go into the form you have

[willow bear]

...x + 2 is the same as 1x + 2

an unwritten coefficient is an implied coefficient of 1

[timid jacinth]

Right but where does your + sign go

[willow bear]

...uh

that's

[timid jacinth]

It looks like they're being multiplied

[willow bear]

a bit of a tough question to answer?

they aren't

they're meant to be separated by a visually significant space

you can put a dummy symbol like ; between them if you want

[timid jacinth]

This is just a really unfamiliar way to me, can you show me the regular messy way?

[willow bear]

yuck

okay sure

[timid jacinth]

Where am I supposed to be learning all of this

I feel like this has all been skipped over and they're just showing me really rigid methods and not the underlying understanding

[viscid thistle]

well this is a precalc concept

so it probably is on your book too if u use one so i'd suggest checking it out as well